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For one fixed POVM repeated on independently prepared copies, the Born rule gives $m$ outcome probabilities with one normalization constraint. An arbitrary mixed qubit density operator has three independent real parameters,The probability map is affine in $\mathbf{r}$. It can be injective on the full Bloch ball only if $m-1\ge 3$. If its rank is smaller, a nonzero undetected direction gives two nearby physical Bloch vectors with the same probabilities. Hence $m\ge 4$ for an informationally complete POVM.

To attain this bound, take the four directionsThey are the vertices of a regular tetrahedron. The tetrahedral qubit POVM has effects $E_i=(I+{\mathbf{n}}_i\cdot \mathbf{\sigma })/4$, each positive and rank one, and ${\sum }_i{E}_i=I$. The observed probabilities areSince ${\sum }_i{\mathbf{n}}_i=0$ and ${\sum }_i{\mathbf{n}}_i{\mathbf{n}}_i^T=4I_3/3$, they recover the state byThis is the minimum for a single fixed measurement family. If the wording permits changing measurement settings between groups of copies, three binary projective measurements of the three Pauli observables also suffice, each with $m=2$. That procedure is informationally complete three-observable qubit tomography, and it does not contradict the four-outcome bound for one fixed POVM. Tomography estimates probabilities statistically from many copies; a finite sample does not determine an arbitrary continuous state exactly.

Quantum state tomography estimates a density operator from measurement frequencies on many independently and identically prepared systems. One fixed projective measurement reveals only its basis populations, so further measurement settings are generally needed. For a qubit, writeThe three Pauli matrices thus turn measured expectation values into the Bloch vector. This reconstruction uses the Bloch ball representation of arbitrary qubit density operators. ^[ref].

Divide the prepared ensemble into groups measured in the $X$, $Y$ and $Z$ bases. For example, positive-outcome frequencies $0.8,0.5,0.9$ give the estimate $\mathbf{r}=(0.6,0,0.8)$ andHere $|\mathbf{r}|=1$, so the reconstructed state is the pure state $(3|0⟩+|1⟩)/\sqrt{10}$. Its $Z$ frequencies alone also fit the diagonal mixed state with populations $0.9,0.1$; the $X$ data distinguish them. For the angular pure state in measurement in quantum mechanics, $r_x=\sin \theta \cos \varphi$, $r_y=\sin \theta \sin \varphi$ and $r_z=\cos \theta$, so $Y$ resolves the sign of the phase that $X$ alone cannot.

Finite samples introduce statistical uncertainty: with $N$ independent binary trials in one basis, the frequency has variance $p(1-p)/N$. At $p=1/2$ and $N=100$, its standard deviation is $0.05$. Fluctuations can place a naive Bloch vector estimate outside the Bloch ball; a physical reconstruction must enforce positivity of the density operator. Repeated measurements on one unprepared specimen sample its changing conditional state, rather than independent copies of its original state. Multiple bases are needed in this projective protocol; a single fixed informationally complete POVM, such as the tetrahedral qubit POVM, is an alternative. It still requires an ensemble of trials.

Choose the four unit Bloch vectors at the vertices of a regular tetrahedron, so ${\sum }_i{\mathbf{n}}_i=0$ and ${\sum }_i{\mathbf{n}}_i{\mathbf{n}}_i^T=4I/3$. The displayed four positive rank-one effects sum to $I$. On input $(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, their probabilities are $p_i=(1+\mathbf{r}\cdot {\mathbf{n}}_i)/4$, giving the reconstruction $\mathbf{r}=3{\sum }_i{p}_i{\mathbf{n}}_i$. This realizes the smallest fixed qubit informationally complete POVM.